Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^3+y^3=8-6xy\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)-8+6xy=0\)
\(\Leftrightarrow\left(x+y\right)^3-2^3-3xy\left(x+y-2\right)=0\)
\(\Leftrightarrow\left(x+y-2\right)\left[\left(x+y\right)^2+2\left(x+y\right)+4\right]-3xy\left(x+y-2\right)=0\)
\(\Leftrightarrow\left(x+y-2\right)\left(x^2+y^2-xy+2x+2y+4\right)=0\)
\(\Leftrightarrow\left(x+y-2\right)\left(2x^2+2y^2-2xy+4x+4y+8\right)=0\)
\(\Leftrightarrow\left(x+y-2\right)\left[\left(x-y\right)^2+\left(x+2\right)^2+\left(y+2\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y-2=0\\\left(x-y\right)^2=\left(x+2\right)^2=\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+y=2\\x=y=-2\left(loại\right)\end{matrix}\right.\)
\(x+y=1\Rightarrow y=1-x\)
\(P=x^3+\left(1-x\right)^3+x\left(1-x\right)\)
\(P=2x^2-2x+1=\dfrac{1}{2}\left(2x-1\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)
\(P_{min}=\dfrac{1}{2}\) khi \(x=y=\dfrac{1}{2}\)
Từ giả thiết ta có:
\(x+y=3\left(\sqrt{x+1}+\sqrt{y+2}\right)\le3\sqrt{2\left(x+y+3\right)}\)
\(\Leftrightarrow P\le3\sqrt{2\left(P+3\right)}\)
\(\Leftrightarrow\left\{{}\begin{matrix}P\ge0\\18P+54\ge P^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}P\ge0\\P^2-18P-54\le0\end{matrix}\right.\)
\(\Leftrightarrow0\le P\le9+3\sqrt{15}\)
\(\Rightarrow maxP=9+3\sqrt{15}\Leftrightarrow\left(x;y\right)=\left(\dfrac{10+3\sqrt{15}}{2};\dfrac{8+3\sqrt{15}}{2}\right)\)
Ta có :
\(x^2+y^2+xy=3\)
\(\Rightarrow\left(x+y\right)^2-xy=3\)
\(\Rightarrow \left(x+y\right)^2=3+xy\)
hay \(S^2=3+xy\le3+\frac{\left(x+y\right)^2}{4}=3+\frac{S^2}{4}\)
\(\Rightarrow S^2\le3+\frac{S^2}{4}\)
\(\Rightarrow S^2\le4\)
\(\Rightarrow-2\le S\le2\)
GTLN của S = 2
x^2+y^2=(x+y)^2-2xy
=5^2-2*3
=25-6
=19
x^3+y^3=(x+y)^3-3xy(x+y)
=5^3-3*3*5
=125-9*5
=80
(x-y)^2=(x+y)^2-4xy=5^2-4*3=13
=>\(x-y=\sqrt{13}\)
